Chapter 3 / Hicksian Demand and Duality

3.2 Cost Minimization with Lagrange

Utility maximization and cost minimization are both constrained optimization problems of the form $\begin{aligned} \max_{x_1,x_2}\ \ \ & f(x_1,x_2) \\ \text{s.t.}\ \ \ & g(x_1,x_2) = 0 \end{aligned}$ In this kind of constrained optimization problem we call the function $f(x_1,x_2)$ the objective function and the equation $g(x_1,x_2) = 0$ the constraint.

Under certain conditions, we may find the solution to a constrained optimization problem by setting up the Lagrangian $\mathcal{L}(x_1,x_2,\lambda) = f(x_1,x_2) + \lambda g(x_1,x_2)$ The first-order conditions set the partial derivatives of this equal to zero, giving us $\begin{aligned} \frac{\partial \mathcal{L}}{\partial x_1} = \frac{\partial f}{\partial x_1} + \lambda \frac{\partial g}{\partial x_1} &= 0 \Rightarrow \lambda = -\frac{\partial f/\partial x_1}{\partial g/\partial x_1}\\ \frac{\partial \mathcal{L}}{\partial x_2} = \frac{\partial f}{\partial x_2} + \lambda \frac{\partial g}{\partial x_2} &= 0 \Rightarrow \lambda = -\frac{\partial f/\partial x_2}{\partial g/\partial x_2}\\ \frac{\partial \mathcal{L}}{\partial \lambda} = g(x_1,x_2) &= 0 \end{aligned}$ Setting equations (1) and (2) equal to one another gives us the “tangency condition” $\frac{\partial f/\partial x_1}{\partial g/\partial x_1} = \frac{\partial f/\partial x_2}{\partial g/\partial x_2}$ or $\frac{\partial f/\partial x_1}{\partial f/\partial x_2} = \frac{\partial g/\partial x_1}{\partial g/\partial x_2}$ So the solution here may be described as “the point along the constraint where the tangency condition is met.”

For what we’ve seen thus far, the objective function has been the “utility function” whose output is measured in utils, and the constraint has been the budget constraint: that is, $\begin{aligned} \textbf{Objective function: }f(x_1,x_2) &= u(x_1,x_2)\\ \textbf{Constraint (when set equal to zero): }g(x_1,x_2) &= m - p_1x_1 - p_2x_2 \end{aligned}$ Mathematically, we can see that $\frac{\partial f}{\partial x_1} = MU_1$ $\frac{\partial f}{\partial x_2} = MU_2$ $\frac{\partial g}{\partial x_1} = -p_1$ $\frac{\partial g}{\partial x_2} = -p_2$ so our values for $\lambda$ in equations (1) and (2) have been $\lambda = -\frac{\partial f/\partial x_1}{\partial g/\partial x_1} = \frac{MU_1}{p_1}$ $\lambda = -\frac{\partial f/\partial x_2}{\partial g/\partial x_2} = \frac{MU_2}{p_2}$ $\frac{\partial f/\partial x_1}{\partial f/\partial x_2} = \frac{\partial g/\partial x_1}{\partial g/\partial x_2} \Rightarrow \frac{MU_1}{MU_2} =\frac{p_1}{p_2}$

For the cost minimization problem, we swap the objective function and the constraint, which is now given by an exogenous level of income $\bar u$: $\begin{aligned} \min_{x_1,x_2}\ \ \ & p_1x_1 + p_2x_2 \\ \text{s.t.}\ \ \ & u(x_1,x_2) = \bar u \end{aligned}$ Now, using the same $f()$ and $g()$ framing as above, we have $\begin{aligned} \textbf{Objective function: }f(x_1,x_2) &= p_1x_1 + p_2x_2\\ \textbf{Constraint (when set equal to zero): }g(x_1,x_2) &= \bar u - u(x_1,x_2) \end{aligned}$

Mathematically, following the same procedure as above, we have $\frac{\partial f}{\partial x_1} = p_1$ $\frac{\partial f}{\partial x_2} = p_2$ $\frac{\partial g}{\partial x_1} = -MU_1$ $\frac{\partial g}{\partial x_2} = -MU_2$ so our values for $\lambda$ in equations (1) and (2) are now $\lambda = -\frac{\partial f/\partial x_1}{\partial g/\partial x_1} = \frac{p_1}{MU_1}$ $\lambda = -\frac{\partial f/\partial x_2}{\partial g/\partial x_2} = \frac{p_2}{MU_2}$ <!– Rather than being “bang for the buck” (measured in utils per dollar), these could be thought of as “buck for the bang” (measured in dollars per util) – that is, $\lambda$ now measures the cost, in dollars, of increasing one’s utility by 1.

Of course, setting the two values of $\lambda$ equal to one another give us the exact same tangency condition as before: –> $\frac{\partial f/\partial x_1}{\partial f/\partial x_2} = \frac{\partial g/\partial x_1}{\partial g/\partial x_2} \Rightarrow \frac{p_1}{p_2} = \frac{MU_1}{MU_2}$ This is the very same tangency point that solves the utility-maximization problem — the duality we explored in the previous reading.

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